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Electrostatic field and potential of an electric dipole inside a conductor

  1. Dec 26, 2009 #1
    1. The problem statement, all variables and given/known data

    Given an electrical dipole of electrical dipole momentum [tex]\vec P = p\hat k[/tex], centered in [tex]0\hat i + 0\hat j + 0\hat k[/tex], find the potential in all the space, where [tex]V(\infty ) = 0[/tex]. If the dipole is now surrounded by a hollow spherical conductor (initially discharged), find the electrostatic field and potential outside the sphere.

    3. The attempt at a solution

    I can find the [tex]\vec E[/tex] and V for all r for the dipole without the sphere: I find that [tex]V(\vec r) = \frac{{\vec P \cdot \vec r}}{{4\pi \varepsilon _0 r^3 }}[/tex]. But the problem comes with the spherical conductor. I can now use Gauss' Law, given that I can use a spherical surface for it. But the net charge inside that surface would be zero (given the dipole and the discharged spherical conductor). This got me thinking that there must be something wrong alltogether, because what I imagine happens is that in one side of the conductor outter surface there would be positive charge and on the other negative charge (due to the presence of the dipole). And that would generate an electrostatic field, similar to that of the dipole.

    What am I doing wrong? Can I use Gauss' Law here?
  2. jcsd
  3. Dec 26, 2009 #2
    A conductor in this context is defined as an equi-potential volume or surface (Assuming equilibrium). This also means that the electric field inside the conductor is 0, but that is a bit more dodgy in this case since we're dealing with an infinitely thin conductor.

    So what's going to happen is that the charges in the conductor arrange themselves so as to provide an equal potential for every point on the conductor.

    How to go on from here, is something I'm not sure on how to do, though..

    Interesting question!
  4. Dec 26, 2009 #3
    Yes. What I'm most baffled about is the fact that I can't use Gauss' Law here. I'd like to believe that the conductor behaves as a big dipole, but I can't find an expression for that.
  5. Dec 26, 2009 #4


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    Don't forget that Gauss's Law still applies... there's just no guarantee that it's going to be useful.

    I think we can use the fact that RoyalCat brought up, that the conductor is an equipotential surface. That sets the boundary conditions for Poisson's equation,
    [tex]\nabla^2 V = 0[/tex]
    in the region outside the sphere, namely that the potential is some constant (say, [itex]V_0[/itex]) at the radius of the sphere. But Poisson's equation is guaranteed to have a unique solution for any given set of boundary conditions, and the same set of boundary conditions (i.e. constant at a given radius) can be produced by a single point charge of appropriate magnitude and no conducting sphere. Thus the potential produced by the dipole inside the conducting shell will be the same as the potential produced by the point charge without the shell. Of course, this argument doesn't tell you the potential inside the sphere, but thankfully the question doesn't ask for that.
  6. Dec 27, 2009 #5
    I think there's something wrong about that. Going back to my notes, I found this problem (a dipole surrounded by a hollow conductor) and it says that outside the conductor E = 0 (it doesn't say why). If that is true, then outside the conductor every r has the same potential.

    Anybody knows if this is right?
  7. Dec 27, 2009 #6


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    I think it is right. The dipole will induce an inhomogeneous charge distribution on the inner surface of the conductor, and the field of this surface charge distribution together with that of the dipole should ensure zero electric field inside the conductor. The total surface charge on the inner surface is zero, that is the same for the outer surface. The electric field inside the conductor is zero, there is nothing to drive redistribution of charge at the outer surface. The metal sphere carries no charge, so the electric field outside it is also zero which means constant potential.

    The situation is similar to the capacitor. There is an electric field between two metal plates with equal and opposite charges, but outside the field is zero.

  8. Dec 27, 2009 #7
    Why should we infer from the fact that there is no charge inside the metal sphere or on it, that the electric field outside it is zero..? Wouldn't that be true only for the volume of the conductor?
  9. Dec 28, 2009 #8


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    It is true that there is zero electric field within the volume of the conductor (i.e. in the volume occupied by the metal), but that doesn't mean that that's the only volume in which there is zero electric field.

    If you know a thing or two about conductors, you might be able to intuitively tell that they shield higher multipole moments - that is, a conducting spherical shell will mask the distribution of charge (but not the total amount of charge) within it. To prove it, though, I think you need to use the argument about uniqueness of solutions to Poisson's equation. I tried to think of a more intuitive (but still convincing) way to reason through it and couldn't come up with anything. Maybe someone else can.
  10. Dec 28, 2009 #9


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    Well, my previous argument should be quite wrong. Now I try two equal and opposite point charges placed symmetrically around the centre inside a hollow metal sphere, and apply the mirror image method but with no success up to now.

  11. Oct 7, 2011 #10
    Hi, this is my first post.
    We recently asked this question to our first year in Physics degrre in Madrid, Spain.

    The answer is very intuitive: Gauss law always can be applied, the problem is that with non uniform Electric field the surface integral cannot be solved. This is not the case when we apply Gauss outside of the spherical conductor: the net charge inside is zero, because the dipole total charge is zero. So outside the spherical conductor there is no Electric field. This is like a Faraday cage or the basic principle of shielding wires: a conducting cover allow no electromagnetic field get inside, if this shield is connected to ground. In this problem there is no ground conection but the net charge is zero, which is the important key.

    Due to the fact that the electric potential is always continuos and is zero outside is zero inside the electric conductor as well.

    The only problem here is the inner surface of the conductor: the free charges of the conductor would redistribute inside but there is no more effect outside. I still have a doubt wether the redistributed charges in the inner surface would recombine locally, but anyway, qualitatively, in the inner surface the electric potential should be zero as well (becase its continuity) and therefore also the electric field. The idea of having an bigger dipole on the outer surface (inversely polarized due to the induction) is only possible if the sphere is non conductive, namely, a dielectric material.

    As far as I know this is the answer.
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